The problem is as follows:
Assume that $V_1$ and $V_2$ are independent random variables with $V_1 \sim \chi^2(5), V_2\sim\chi^2(9)$.
Find the value of $b$ such that:
$$P[\frac {V_1}{V_1 + V_2} \lt b] = 0.90$$
What I've tried:
$V_1 + V_2 \sim \chi^2(5+9) $
$P[\frac {V_1}{V_1 + V_2} \lt b] = P[\frac {V_1/5}{(V_1 + V_2)/14} \lt \frac{14b}{5}]$
We know that $\frac{V_1/5}{(V_1 + V_2)/14}\sim F(5,14)$
Then I used a F distribution calculator to obtain that $\frac{14b}{5}=2.31$
Hence $b = 0.825$
However, the answer in the book gives $b = 0.592$
What did I do wrong here?
The error is that you have have assumed that the distribution of $\frac{V_1}{V_1+V_2}$ is distributed as $F(5,14)$. This is not the case, as $V_1$ and $V_1+V_2$ are not independent (have a look at the Characterisation section in here) - on the other hand, $\frac{V_1/5}{V_2/9}$ would be distributed as $F(5,9)$.
Going back to the problem at hand, we can tackle it as follows:- $$\begin{align}P\left[\frac{V_1}{V_1+V_2}<b\right]&=P\left[1+\frac{V_2}{V_1}>\frac{1}{b}\right]\\&=P\left[\frac{V_2}{V_1}>\frac{1}{b}-1\right] \\&=P\left[\frac{V_2/9}{V_1/5}>\frac{5}{9}\left(\frac{1}{b}-1\right)\right]\\&=1-P\left[\frac{V_2/9}{V_1/5}\leq\frac{5}{9}\left(\frac{1}{b}-1\right)\right]=0.9\end{align}$$
So we want to find the quantile of $F(9,5)$ for $0.1$ - from F-tables, it turns out to be about $0.3831$, so that $b$ is $$b=\frac{1}{1+(0.3831\times(9/5))}=0.592$$